# Acetylene gas (C2H2) is used in welding. During one job, a welder burns 48.0 g of acetylene. Using a balanced equation for the complete combustion of acetylene, calculate the following:

What is the balanced equation.
The mass of oxygen used?
The mass of carbon dioxide produced?
The mass of water produced?

## To determine the information required, we'll first need to write the balanced equation for the complete combustion of acetylene (C2H2).

The balanced equation for the combustion of acetylene is:
C2H2 + 2.5 O2 → 2 CO2 + H2O

Now let's calculate the mass of oxygen used, mass of carbon dioxide produced, and the mass of water produced.

1. Mass of oxygen used:
Looking at the balanced equation, we can see that 1 mole of acetylene (C2H2) reacts with 2.5 moles of oxygen (O2). To calculate the mass of oxygen used, we need to convert the mass of acetylene burned (48.0 g) into moles.

The molar mass of acetylene (C2H2) is:
(2 × atomic mass of carbon) + (2 × atomic mass of hydrogen)
= (2 × 12.01 g/mol) + (2 × 1.01 g/mol)
= 26.04 g/mol

Using the molar mass of acetylene (C2H2), we can find the number of moles burned:
Number of moles = Mass / Molar mass
Number of moles = 48.0 g / 26.04 g/mol
Number of moles ≈ 1.84 mol

According to the balanced equation, 1 mole of acetylene reacts with 2.5 moles of oxygen. Therefore, the mass of oxygen used can be calculated as:
Mass of oxygen used = Number of moles (acetylene) × Molar mass (oxygen)
Mass of oxygen used = 1.84 mol × (2.5 × 32.00 g/mol)
Mass of oxygen used ≈ 118 g

Therefore, the mass of oxygen used is approximately 118 grams.

2. Mass of carbon dioxide produced:
From the balanced equation, we can see that 1 mole of acetylene (C2H2) produces 2 moles of carbon dioxide (CO2). To calculate the mass of carbon dioxide produced, we can use the number of moles of acetylene burned.

The molar mass of carbon dioxide (CO2) is:
(1 × atomic mass of carbon) + (2 × atomic mass of oxygen)
= 12.01 g/mol + (2 × 16.00 g/mol)
= 44.01 g/mol

Using this molar mass, we can calculate the mass of carbon dioxide produced:
Mass of carbon dioxide produced = Number of moles (acetylene) × Molar mass (carbon dioxide)
Mass of carbon dioxide produced = 1.84 mol × (2 × 44.01 g/mol)
Mass of carbon dioxide produced ≈ 161.8 g

Therefore, the mass of carbon dioxide produced is approximately 161.8 grams.

3. Mass of water produced:
The balanced equation shows that 1 mole of acetylene (C2H2) produces 1 mole of water (H2O). Hence, to calculate the mass of water produced, we can use the number of moles of acetylene burned.

The molar mass of water (H2O) is:
(2 × atomic mass of hydrogen) + (1 × atomic mass of oxygen)
= (2 × 1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol

Using this molar mass, we can calculate the mass of water produced:
Mass of water produced = Number of moles (acetylene) × Molar mass (water)
Mass of water produced = 1.84 mol × 18.02 g/mol
Mass of water produced ≈ 33.10 g

Therefore, the mass of water produced is approximately 33.10 grams.

To summarize:
- The balanced equation for the complete combustion of acetylene is:
C2H2 + 2.5 O2 → 2 CO2 + H2O

- The mass of oxygen used is approximately 118 grams.
- The mass of carbon dioxide produced is approximately 161.8 grams.
- The mass of water produced is approximately 33.10 grams.

## To answer the questions, let's first write the balanced chemical equation for the complete combustion of acetylene (C2H2):

C2H2 + 5/2 O2 -> 2 CO2 + H2O

From the balanced equation, we can see that 1 mole of C2H2 reacts with 5/2 moles of O2 to produce 2 moles of CO2 and 1 mole of H2O.

Now let's calculate the answers step-by-step:

1. The mass of oxygen used:
From the equation, we know that the molar ratio between C2H2 and O2 is 1:5/2. To calculate the mass of O2 used, we need to convert the mass of acetylene to moles, and then use the molar ratio to determine the moles of O2 used, and finally convert the moles of O2 used to grams.

Molar mass of C2H2 = 2 * atomic mass of Carbon + 2 * atomic mass of Hydrogen
= 2 * 12.01 g/mol + 2 * 1.008 g/mol
= 26.04 g/mol

Moles of C2H2 = Mass of C2H2 / Molar mass of C2H2
= 48.0 g / 26.04 g/mol
≈ 1.842 mol

Moles of O2 = Moles of C2H2 * (5/2 moles of O2 / 1 mole of C2H2)
= 1.842 mol * (5/2)
= 4.605 mol

Mass of O2 = Moles of O2 * Molar mass of O2
= 4.605 mol * (2 * 16.00 g/mol)
= 147.36 g

Therefore, the mass of oxygen used is approximately 147.36 grams.

2. The mass of carbon dioxide produced:
From the equation, we know that the molar ratio between C2H2 and CO2 is 1:2. To calculate the mass of CO2 produced, we use the same approach as above.

Moles of CO2 = Moles of C2H2 * (2 moles of CO2 / 1 mole of C2H2)
= 1.842 mol * 2
= 3.684 mol

Mass of CO2 = Moles of CO2 * Molar mass of CO2
= 3.684 mol * (12.01 g/mol + 2 * 16.00 g/mol)
= 132.92 g

Therefore, the mass of carbon dioxide produced is approximately 132.92 grams.

3. The mass of water produced:
From the equation, we know that the molar ratio between C2H2 and H2O is 1:1. To calculate the mass of H2O produced, we use the same approach as above.

Moles of H2O = Moles of C2H2 * (1 mole of H2O / 1 mole of C2H2)
= 1.842 mol

Mass of H2O = Moles of H2O * Molar mass of H2O
= 1.842 mol * (2 * 1.008 g/mol + 16.00 g/mol)
= 58.09 g

Therefore, the mass of water produced is approximately 58.09 grams.